如題（大數乘法）。把FFT種的n次單位根換成模意義下的“n次單位原根”，即$G^{\frac{p-1}{n}}$即可。

//#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
//#include<math.h>
//#include<time.h>
//#include<complex>
#include<algorithm>
using namespace std;

int n,m,wei;
#define maxn 262222
const
 
 int mod=998244353,G=3;
int a[maxn],b[maxn],c[maxn],rev[maxn];

int powmod(int a,int b)
{
    int ans=1;
    while (b)
    {
        if (b&1) ans=1ll*ans*a%mod;
        a=1ll*a*a%mod;
        b>>=1;
    }
    return ans;
}

void dft(int *a,int n,int type)
{
    for (int i=0;i<n;i++) if (i<rev[i]) {int
 
 t=a[i]; a[i]=a[rev[i]]; a[rev[i]]=t;}
    for (int i=1;i<n;i<<=1)
    {
        int w=powmod(G,(mod-1)/(i*2));
        if (type==-1) w=powmod(w,mod-2);
        for (int j=0,p=i<<1;j<n;j+=p)
        {
            int t=1;
            for (int k=0;k<i;k++,t=1ll*t*w%mod)
            {
                
 
int tmp=a[j+k+i]*1ll*t%mod;
                a[j+k+i]=(a[j+k]-tmp+mod)%mod;
                a[j+k]=(a[j+k]+tmp)%mod;
            }
        }
    }
}

void ntt(int *a,int *b,int *c)
{
    if (!rev[1]) for (int i=0;i<n;i++)
    {
        rev[i]=0;
        for (int j=0;j<wei;j++) rev[i]|=((i>>j)&1)<<(wei-j-1);
    }
//    for (int i=0;i<n;i++) cout<<rev[i]<<‘ ‘;cout<<endl;
    dft(a,n,1); dft(b,n,1);
    for (int i=0;i<n;i++) c[i]=1ll*a[i]*b[i]%mod;
    dft(c,n,-1);
    int inv=powmod(n,mod-2);
    for (int i=0;i<n;i++) c[i]=1ll*inv*c[i]%mod;
}

char sa[maxn],sb[maxn];
int main()
{
    scanf("%s%s",sa,sb); n=strlen(sa)-1; m=strlen(sb)-1;
    for (int i=0;i<=n;i++) a[i]=sa[n-i]-‘0‘;
    for (int i=0;i<=m;i++) b[i]=sb[m-i]-‘0‘;
    m=m+n; for (n=1,wei=0;n<=m;n<<=1,wei++);
    ntt(a,b,c);
    for (int i=0;i<=m;i++) c[i+1]+=c[i]/10,c[i]%=10;
    if (c[m+1]) m++;
    for (int i=m;i>=0;i--) printf("%d",c[i]);
    return 0;
}

NTT模板